13 CHEMICAL EQUILIBRIUM W MODULE - 5

MODULE - 5
Chemistry
Chemical Dynamics
13
CHEMICAL EQUILIBRIUM
Notes
hen reactants are mixed in exact stoichiometric proportion to perform a chemical
W
reaction, it is believed that all the reactants would be converted into products with the
release or absorption of energy. This is not true in all cases. Many chemical reactions
proceed only to a certain extent and stop. When analysed, the resulting mixture contains
both the reactants and products. It is because when reactants combine to form products,
the products also start combining to give back the reactants.
When such opposing processes take place at equal rates, no reaction appears to take
place and it is said that a state of equilibrium has reached. In this lesson, we will examine
many aspects of chemical equilibrium. We shall also discuss how can we control the
extent to which a reaction can proceed by changing the various conditions of the equilibrium.
Objective
After reading this lesson you will able to :
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
differentiate between static and dynamic equilibrium;

identify and differentiate between reversible and irreversible reactions;

explain the reversible reaction occuring at the equilibrium state;

list and explain characteristics of equilibrium state;

apply the law of equilibrium and write expression of equilibrium constant for different
types of equilibria, namely physical, chemical, homogeneous and heterogenous;

state and derive the relation between Kc and Kp and carry out some calculations
involving them and

list the factors which affect the state of equilibrium and state and apply
Le-Chatelier principle.
Chemical Equilibrium
13.1 Static and Dynamic Equilibrium
The state equilibrium can be observed in physical and chemical systems. Also, equilibrium
can be static or dynamic in nature. A book lying on the table is an example of static
equilibrium. The forces of action and reaction cancel each other and no change takes
place. Thus it is a case of static equilibrium. On the other hand, when an escalator is
coming down and a passenger is going up at the same speed it is a case of dynamic
equilibrium. Here, because both are moving in opposite directions and at the same speed,
no net change takes place. The equilibrium established in the above examples are in
physical systems.
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Chemical Dynamics
Notes
13.2 Reversible and Irreversible Reactions
Chemical reactions can be classified as : Reversible and Irreversible reactions.
13.2.1 Reversible reactions
Consider the reaction between ethanol and acetic acid. When mixed in the presence of
dilute sulphuric acid they react and form ethyl acetate and water.
H
C2H5OH(l) + CH3COOH (l) 
 CH3COO C2H5(l) + H2O(l)
On the other hand, when ethyl acetate and water are mixed in the presence of dilute
sulphuric acid the reverse reaction occurs.
H
CH3COOC2H5(l) + H2O(l) 
 CH3COOH(l) + C2H5OH(l)
It may be noted here that the second reaction is reverse of the first one and under the
same conditions, the two reactions occur simultaneously. Such reactions which occur
simultaneously in opposite directions are called reversible reactions.
A reaction is said to be reversible if under certain conditions of temperature and
pressure, the forward and reverse reactions occur simultaneously.
Reversible reactions are indicated by placing two half arrows pointing in opposite directions
(  ) between the reactants and products. Thus the above reaction is more appropriately
written as
CH3COOH (l) + C2H5OH (l)  CH3COOC2H5 (l) + H2O(l)
When ethyl acetate and water are formed in the forward reaction the reverse reaction
also starts in which ethanol and acetic acid are formed. After some time the concentrations
of all the reactants and products become constant. This happens when the rates of forward
and reverse reactions become equal; and all the properties of the system become constant.
It is said that the system has attained state of equilibration. However it may be noted
that the state of equilibrium is reached only if the reaction is carried out in a closed
system. At the time of equilibrium, forward and reverse reactions are taking place and it
is in a state of dynamic equilibrium because no change is taking place.
A reversible reaction is said to be in the equilibrium state when the forward and
backward reaction occur simultaneously at the same rate in a closed system and
the concentrations of reactants and products do not change with time
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A common example of reversible reactions of the type A + B  C + D
 CH3COOH + H2O
CH3COOH + C2H5OH 
The following graphs Fig. 13.1 shows the equilibrium state in a reversible reaction.
Notes
Concentration
C or D
A or B
Time
Equilibrium
Fig. 13.1 : Equilibrium in reversible reaction
The graph depicts that the rate of forward reaction gradually decreases while the rate of
backward reaction increase till they become constant and equal to each other.
13.2.2 Irreversible Reactions
Most of the reactions occur only in one direction. They are called irreversible reactions.
For example when carbon is burnt in air to form carbon dioxide the reaction goes only in
one direction i.e. in the direction of formation of carbon dioxide
C (s) + O2 (g) 
 CO2 (g)
Strictly speaking all reactions are considered to be reversible. But the rate of reaction in
one particular direction is extremely small compared to the other. Thus the reaction
proceeds practically in one direction to near completion, leaving a negligibly small amount
of reactant at the end.
When hydrochloric acid is mixed with sodium hydroxide, a base, in equimolar quantities,
a neutralisation reaction takes place; with the formation of salt and water.
HCl (aq) + NaOH (aq) 
 NaCl (aq) + H2O (l)
This reaction proceeds to completion in the forward direction. Similarly when a solution
of silver nitrate is added to a solution of sodium chloride silver chloride is precipitated
immediately.
NaCl (aq) + AgNO3 (aq) 
 AgCl (s) + NaNO3 (aq)
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Chemical Equilibrium
13.3 Characteristics of Equilibrium State
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Chemical Dynamics
1. The state of chemical equilibrium is reached in a reversible reaction when;
(i) the temperature of the system attains a constant value.
(ii) the pressure of the system attains a constant value.
(iii) the concentrations of all the reactants and products attain constant values.
Notes
The state of equilibrium has following characteristics properties :
(i) Chemical Equilibrium is dynamic in nature
The chemical equalibrium is the result of two equal but opposite processes occurring in
the forward and reverse directions and there is no “net” change occurring in the system.
(ii) Equilibrium can be attained from either side
The same state of equilibrium (characterized by its equilibrium constant which is discussed
later can be reached whether the reaction is started from the reactants or products side.
For example, the same equilibrium
 2NO2 (g)
N2O4 (g) 
is established whether we start the reaction with N2O4 or NO2.
(iii) Equilibrium can be attained only in a closed system
Equilibrium can be attained only if no substance among, reactants or products, is allowed
to escape i.e. the system is a closed one. Any system consisting of gaseous phase or
volatile liquids must be kept in a closed container, e.g.
N2(g) + 3H2(g)  2NH3(g)
A system consisting of only non-volatile liquid and solid phases can be kept even in an
open container because such substances have no tendency to escape, e.g.
FeCl3(aq) + 3 NH4SCN(aq)  Fe (SCN)3 (s) + 3 NH4Cl(aq)
(iv) A catalyst can not change the equilibrium state
Addition of a catalyst speeds up the forward and reverse reactions by same extent and
help in attaining the equilibrium faster. However, the equilibrium concentrations of reactants
and products are not affected in any manner.
13.4 Equilibrium in Physical Processes; Phase Equilibrium
State of equilibrium can also be reached in physical processes.
13.4.1 Liquid – Vapour Equilibrium
Let us take some quantity of a liquid in an empty container and close it. Initially the vapour
pressure above the liquid will be zero. The liquid will evaporate and its vapour will fill the
empty space above it.
Liquid  Vapour
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Chemistry
The rate of evaporation is maximum in beginning. As vapours build up, its pressure increases
and the rate of evaporation slows down. Also the reverse process of condensation begins
(Fig. 13.2).
Vapour  Liquid
Vapour
Notes
Liquid
Fig. 13.2 : Liquid Vapour equilibrium
and its rate gradually increases with the increase in the vapour pressure. After some time
the two rates (of evaporation and condensation) become equal and the following equilibrium
is established.
Liquid  Vapour
At equilibrium the vapour pressure reaches its maximum value and is known as the
saturated vapour pressure or simply the vapour pressure. At a fixed temperature, each
liquid has its own characteristic vapour pressure. The vapour pressure of a liquid increases
with rise in temprature.
13.4.2 Solid – Vapour Equilibrium
Volatile solids sublime form vapour. The situation is just similar to the liquid vapour system.
When kept in a closed container at a constant temperature the following equilibrium is
established.
Solid  Vapour
Iodine
Vapour
Solid
Iodine
Fig. 13.3 : Solid vapour equilibrium
Such an equilibrium can be established by keeping some solid iodine in a gas jar covered
with a lid. (Fig. 13.3). Gradually the purple coloured iodine vapours fill the jar and the
following equilibrium is established.
I2(s)  I2(g)
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Chemical Equilibrium
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Chemical Dynamics
13.4.3 Solid – Liquid Equilibrium
Below its freezing point a liquid freezes spontaneously
Liquid  Solid
When heated above its melting point the solid melts spontaneously :
Solid  Liquid
Notes
At the melting point, the two phases are in equilibrium
Solid  Liquid
because the above two processes occur simultaneously and at the same rate. Such an
equilibrium is characterized by its temperature i.e. the melting point of the solid.
13.4.4 Solute – Solution Equilibria
Saturated
Solution
Undissdued
Sugar
Fig. 13.4 : Solute - Solution Equilibrium
When sugar crystals are put in a saturated solution of sugar in water; it will appear that no
change is taking place and sugar appears to remain undissolved. Actually, the undissolved
sugar does dissolve in the saturated sugar solution; and an equal amount of sugar seperates
out from the solution. The solid sugar and the sugar solution form an equilibrium system
which is dynamic in nature.
sugar (s)  sugar solution (saturated)
The equilibrium is established when the rate of dissolution of sugar becomes equal to the
rate of crystallisation. In general such equilibrium can be represented as
solute (s)  solution (saturated)
This equilibrium is known as Solubility Equilibrium.
13.4.5 Phase and Phase Equilibrium
You must have noticed in each of the above equilibria the system consists of two distinct
parts; solid, liquid, solution or vapour. Each of these parts is called a phase.
A phase is defined as a homogenous part of a system which has uniform
composition and properties throughout.
A phase is not the same as physical state. A mixture of two solids, even when powdered
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Chemical Dynamics
Notes
Chemistry
finely is a two phase system. This is because particles of the two solids have different
chemical compositions and physical properties. Completely miscible liquids, solutions and
all gaseous mixture constitute only one phase each.
All the cases of physical equilibrium are in fact the systems in which different phases are
in equilibrium; only if they contain, at least one common component. A dynamic exchange
of the common component between two phases takes place. When the rates of exchange
becomes equal the equilibrium is established. In solid solute and solution equilbrium the
example given earlier, sugar is the common component.
13.5 Equilibrium in Homogeneous and Heterogeneous Systems
13.5.1 Homogeneous and Heterogeneous Systems
Homogeneous system is one which has one phase. It has the same chemical composition
and uniform properties throughout. It is formed by particles of molecular size only. Pure
solids, liquids, gases and solutions are the examples of homogeneous systems.
A system consisting of only one phase is called a homogeneous system
Heterogeneous system, on the other hand has at least two phases – a mixture of solids or
immiscible liquids constitutes a heterogeneous system.
Any system consisting of two or more phases is called heterogeneous system
13.5.2 Homogeneous and Heterogeneous Equilibrium Systems
Equilibrium can be established in either type of systems. Since all physical equilibria involve
at least two phases, therefore these are all examples of heterogeneous equilibrium. But
chemical equilibrium can be homogeneous or heterogeneous in nature. It is homogeneous
if both the reactants and products are present only in one phase gas or liquid and
heterogeneous if present in more than one phase. In the following sections we shall study
such systems.
13.5.3 Homogeneous Chemical Equilibrium System
(a) Gas – Phase homogeneous systems
Such systems contain only gaseous reactants and products. Since all gaseous mixtures
are homogeneous in nature they constitute only one phase. Following are examples of this
type of equilibrium:
(i) N2 (g) + 3H2 (g)  2NH3 (g)
(ii) 2N2O5 (g)  4NO2 (g) + O2 (g)
(b) Liquid – Phase homogeneous systems
These are the systems in which both the reactants and products are present in only one
liquid phase (as a solution) for example :

H


(i) CH3 COOH (l) + C2H5OH (l) 
 CH3COOC2H5(l) + H2O (l)
 HCN (aq) + KOH (aq)
(ii) KCN (aq) + H2O (l) 
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Chemical Equilibrium
13.5.4 Heterogeneous Chemical Equilibrium Systems
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Chemical Dynamics
The systems in which reactants and products are present in more than one phase belong
to this type. For example :
 Fe O (s) + 4H (g)
(i) Fe (s) + 4H2O (g) 
3 4
2
 CaO (s) + CO (g)
(ii) CaCO3 (s) 
2
Notes
Intext Questions 13.1
1.
What is a reversible reaction? Give two examples.
................................................................................................................................
2.
When does a reaction reach equilibrium state?
................................................................................................................................
3.
How would you know whether a system has reached the equilibrium state or not?
................................................................................................................................
4.
Give two examples of physical equilibrium.
................................................................................................................................
5.
Give two example each of chemical homogeneous and heterogeneous equilibria.
................................................................................................................................
13.6 Quantitative Aspect of Equilibrium State
13.6.1 Law of Equilibrium and Concentration
Equilibrium Constant
Consider the following equilibrium
 2HI (g)
H2(g) + I2(g) 
At equilibrium the concentrations of H2, I2 and HI become constant. Also, it has been
found experimentally that irrespective of the starting concentrations of H2 and I2 the
following ratio of concentration terms always remains constant.
Kc 
[HI]2
[H 2 ][I 2 ]
Here [H2], [I2] and [HI] represent the equilibrium molar concentrations of H2, I2 and HI
respectively and Ke is called the concentration equilibrium constant (some times it is
written as simply K). In general, for reversible reaction
 cC + dD
aA + bB 
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at equilibrium, the following ratio of concentration terms always remains constant at a
given temperature.
Kc 
Notes
[C]c [D]d
[A]a [B]b
The above relation is known as the law of equilibrium. It may be noted here that all the
concentrations values in the law of equilibrium are the equilibrium concentrations of
reactants and products. The numerator of the law of equilibrium is the product of equilibrium
molar concentrations of products, each term being raised to the power equal to its
stoichiometric coefficient in the chemical equation and the denominator contains products
of similar concentration terms of reactants.
13.6.2 Pressure Equilibrium Constant Kp
In case of gases their partial pressures can also be used in place of molar concentrations
(since the two are directly proportional to each other) in the law of equilibrium. The new
equilibrium constant, Kp, is called the pressure equilibrium constant. For the reaction
between H2 and I2, Kp is given by
Kp 
p 2 HI
p H 2  p I2
Here p H2 , p I2 and p HI are the equilibrium partial pressures of H 2, I2 and HI
respectively. For the general gas phase reaction :
a A (g) + b B (g)  c C (g) + d D (g)
it is given by :
kp 
p cC  p dD
p aA  p Bb
13.6.3 Relation between Kp and Kc
For a general gas phase reaction at equilibrium
a A (g) + b B (g)  c C (g) + d D (g)
The pressure and concentration equilibrium constants Kp and Kc are
kp 
p cC  p dD
[C]c [D]d
k

and c
p aA  p Bb
[A]a [B]b
For a gaseous substance i, the ideal gas equation is
piV = niRT
where pi and ni are its partial pressure and amount in a gaseous mixture and V and T are
its volume and temperature and R is the gas constant. The relation may be written as
pi =
234
ni
RT = ci RT
T
V
Chemical Equilibrium
Where ci is the molar concentration or molarity of ‘i’ expressed in moles per litre. This
relation can be used for replacing the partial pressure terms in the expression for K p.
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Chemical Dynamics
(cC RT)c (c D RT)d
Kp =
(c A RT)a (c B RT) b
=
ccC cdD
(RT)
caA c Bb
(c + d) – (a + b)
Notes
Using the square bracket notation for molar concentration the relation can be written as
Kp =
[C]c [D]d
(RT) (n P – n R )
a
b
[A] [B]
= K c (RT)
n g
where ng is the change in the moles of gaseous substances in the reaction and is equal to
the difference in the moles of gaseous products nP and the moles of gaseous reactants, nR.
ng may be zero positive or negative.
(i) In the reaction
H2 (g) + I2 (g)  2HI (g)
Here nP = moles of the gaseous product is equal to 2
nR = moles of gaseous reactant H2 and I2 is equal to 2 (as 1 + 1).
Hence  ng = nP – nR = 2 – 2 = 0
ng = 0
(ii) In the reaction
N2 (g) + 3H2 (g)  2NH3 (g)
nP = 2, nR = 1 + 3 = 4
and  ng = 2 – 4 = – 2
(iii) In the reaction involving solids and gases
CaCO3 (s)  CaO (s) + CO2 (g)
ng = 1
13.6.4 Expressions of Equilibrium Constant for Some Reactions
The law of equilibrium can be applied to write down expressions of K c and Kp for some
reactions
13.7 Homogeneous Equilibria
(i) Decomposition of N2O4
N2O4 (g)  2 NO2 (g)
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Kc 
p 2NO2
[NO 2 ]2
K

; Kp = p p
[N 2 O 4 ]
N 2 O4
(ii) Oxidation of sulphur dioxide
2SO2 (g) + O2 (g)  2SO3 (g)
Notes
2
pSO3
[SO3 ]2
Kc =
; Kp = p 2 .p
2
[SO 2 ] [O 2 ]
SO 2
O2
(iii) Esterification of acetic acid with ethanol
CH3COOH (I) + C2H5OH (I)  CH3COOC2H5 (I) + H2O(l)
[CH 3COOC2 H 5 ][H 2 O]
Kc = [CH COOH][C H OH]
3
2 5
In this reaction no gas is involved, therefore expression for Kp is meaningless.
13.7.1 Heterogeneous Equilibrium
Consider the following equilibrium
CaCO3 (s)  CaO (s) + CO2 (g)
According to the law of equilibrium
Kc =
[CaO][CO2 ]
[CaCO3 ]
Here CaCO3 and CaO are pure solids. The concentration of any solid is constant at a
fixed temperature therefore these are not written in expression for equilibrium constant
for hetrogenous reactions. Equilibrium constants for the reaction can be written as
Kc = [CO2] and Kp = Pco2
Following are some more examples of heterogenous equilibrium
(i) Reaction between iron and steam
3 Fe (s) + 4H2O (g)  Fe3O4 (s) + 4H2 (g)
p 4H2
[H 2 ]4
Kc =
; Kp = p 4
[H 2 O]4
H2O
(ii) Liquid – Vapour Equilibrium
H2O(I)  H2O (g)
Kc = [H2O; g] ; Kp = p H2O .
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Chemical Equilibrium
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13.8 Characteristics of Equilibrium Constant
13.8.1 Equilibrium Constant and Chemical Equation
The expression of equilibrium constant depends upon the manner in which the chemical
equation representing it is written. For the reaction
H2 (g) + I2 (g)  2HI (g)
Notes
[HI]2
The equilibrium constant K is given by K =
[H 2 ][I 2 ]
When the same reaction is written as
(a)
1
1
H2 (g) +
 (g)  HI (g)
2
2 
the corresponding equilibrium constant K1 is given by
[HI]
K1 =
1
1
[H 2 ] 2 [I 2 ] 2
It may be noted that equilibrium constants K and K1 are related as K1 =
K
(b) When the reaction is written as reverse
2HI (g)  H2 (g) + I2 (g)
K2 =
[H 2 ][I 2 ]
[HI]2
Here it can be seen that
K2 =
1
K
Similar relationship is also observed in the pressure equilibrium constant K p. Thus the
expression of equilibrium constant depends on how the reaction is expressed in the form
of a chemical equation.
13.8.2 Units of Equilibrium Constant
Units of equilibrium constant Kc or Kp depend upon the fact whether during the reactions
there is any change in the moles of substance or not.
(a) The reactions in which there is no change in moles of substance i.e.  n = 0.
The equilibrium constant for such reaction has no units. For example in the reaction
between H2 and I2
H2 (g) + I2 (g)  2HI (g)
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Notes
Kc =
[HI]2
[H 2 ][I 2 ]
p 2HI
Kp = p . p
H2
I2
Kc =
(mol L1 ) 2
(mol L1 )(mol L1 )
Kp =
bar 2
(bar)(bar)
 Hence Kp and Kc have no units in such cases.
(b) The reaction where there is change in the moles of substance i.e.  n  0.
The equilibrium constant for such reactions has units which depend upon the change in
moles of substances.
For example :
N2 (g) + 3H2 (g)  2NH3 (g)
 n =  np   nR
=2–4=–2
The units of Kc for this reaction would be (mol L–1)–2 or L2 mol–2 and those of Kp would be
bar–2 as shown below :
The equilibrium constant for such reactions are
[NH 3 ]2
Kc =
[N 2 ][H 2 ]3
Kp = p  p 3
N2
H2
(mol L1 ) 2
Kc =
(mol L1 )(mol L1 )3
= (mol L–1)–2
= L2 mol–2
pressure 2
Kp =
pressure. pressure3
= pressure–2
= bar – 2
For the reaction PCl5 (g)  PCl3 (g) + Cl2 (g)
n = 2 – 1 = 1. Therefore,
The units for Kc and Kp are
Kc = mol L–1 and Kp = bar
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p 2NH3
Chemical Equilibrium
13.8.3 Significance of the Magnitude of K
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Chemical Dynamics
The equilibrium constant of a reaction has a constant and characteristic value at a given
temperature. The changes in starting concentration, pressure and the presence of a catalyst
do not change the value of the equilibrium constant. However if the temperature is changed.
The value of the equilibrium constant also changes.
The magnitude of the equilibrium constant is a measure of the extent upto which a reaction
proceeds before the equilibrium is reached. The magnitude of K is large when the products
are present in larger amounts than the reactants in the equilibrium mixture. For the reaction
H2 (g) + I2 (g)  2 HI (g)
Kc = 90 at 298 K
and for 2CO (g) + O2 (g)  2 CO2 (g)
Kc = 2.2 × 1022 at 1000 K.
Notes
A large value of Kc for the second reaction indicates that amount of products is much
more than the reactants present at the time of equilibrium. Thus the magnitude of equilibrium
constant tells us about the position of the equilibrium.
13.8.4 Calculation of Equilibrium Constants
Equilibrium constants Kc and Kp can be calculated if the equilibrium concentrations or
partial pressures are known or can be obtained from the given data. The following examples
illustrate the calculations.
Example 13.1 : Calculate the equilibrium constant for the reaction
A (g) + B (g)  C (g) + D (g)
If at equilibrium 1 mol of A, 0.5 mole of B, 3.0 mole of C and 10 mol of D are present in
a one litre vessel.
Solution : From the law of equilibrium
[C][D]
Kc = [A][B]
Since the volume of the vessel is one litre, the number of moles of A, B, C and D are equal
to their concentrations. Thus
[A] = 1 mol L–1, [B] = 0.5 mol L–1, [C] = 3.0 mol L–1 and [D] = 10 mol L–1 and
Kc
(3.0 mol L1 ) (10 mol L1 )
=
(1 mol L1 ) (0.5 mol L1 )
3.0  10
= 1  0.5 = 60
Example 13.2 In an experiment carried out at 298 K, 4.0 mol of NOCl were placed in
a 2 litre flask and after the equilibrium was reached 1.32 mol of NO were formed. Calculate
Kc at 298 K for the reaction
2NOCl (g)  2 NO (g) + Cl2 (g)
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Solution Calculation of equilibrium concentrations
(i) [NO] =
Notes
No. of moles of NO 1.32 mol
=
= 0.66 mol L–1
Volume
2L
1
(No. of moles of NO) 1.32 mol
No. of moles of Cl2
(ii) [Cl2] =
= 2
= 2  2L = 0.33 mol L–1
Volume
Volume
(iii) [NOCl] =
=
No. of moles of NOCl (Initial moles - moles decomposed)
=
Volume
Volume
(4.0 – 1.32) mol
2.68 mol
=
= 1.34 mol L–1
2L
2L
For the reaction
2NOCl (g)  2NO (g) + Cl2 (g)
(0.66 mol L1 ) 2 (0.33 mol L1 )
(0.66) 2  0.33
[NO]2 [Cl2 ]
Kc =
=
=
(1.34 mol L1 ) 2
(1.34) 2
[NOCl]2
= 0.080 mol L–1
Kc = 0.080 mol L–1
Example 13.3 : 2 moles of HI were heated in a vessel of one litre capacity at 713 K
till the equilibrium was reached. At equilibrium HI was found to be 25% dissociated.
Calculated Kc and Kp for the reaction.
Solution Initial moles of HI = 2
Moles of HI dissociated =
25  2
= 0.5 mol
100
Moles of HI at equilibrium = 2.0 – 0.5 = 1.5 mol
The dissociation of HI occurs as
2HI (g)
H2 (g)
+
I2 (g)
Initial moles
2
0
0
Equilibrium moles
(2 – 0.5)
0.25
0.25
1.5 mol
0.25 mol
0.25 mol
Volume of reaction vessel
1L
1L
1L
Equilibrium concentration
1.5 mol L–1
0.25 mol L–1
0.25 mol L–1
For the reaction
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
Chemical Equilibrium
MODULE - 5
Chemical Dynamics
(0.25 mol L1 ) (0.25 mol L1 )
[H 2 ][I 2 ]
Kc =
=
(1.5 mol L1 ) 2
[HI]2
(0.25)2
=
= 0.028
(1.5)2
Also Kp = Kc (RT) n g
For this reaction ng = np – nR = 2 – 2 = 0
Notes
 Kp = Kc = 0.028
Example 13.4 : Calculate Kp for the reaction COCl2  CO + Cl2 in atm and Nm–2.
The equilibrium partial pressures of COCl2, CO and Cl2 are 0.20, 0.16 and 0.26 atm
respectively.
(1 atm = 101300 Nm–2)
Solution : (i) Kp in atmospheres
COCl2(g)  CO (g) + Cl2 (g)
Kp =
pco  pCl2
pCOCl2
=
(0.16 atm)(0.26 atm)
0.16  0.26
=
atm
(0.20 atm)
0.20
= 0.21 atm.
(ii) Kp in Nm–2
Kp = 0.21 atm and 1 atm = 101300 Nm–2
 Kp = (0.21 atm) (101300 Nm–2 atm–1) = 21273 Nm–2
Example 13.5 : When equal number of moles of ethanol and acetic acid were mixed at
300 K, two-third of each had reacted when the equilibrium was reached. What is the
equilibrium constant for the reaction?
CH3COOH (l) + C2H5OH (l)  CH3COOC2H5 (l) + H2O (l)
Solution : Let n moles each of acetic acid and ethanol be mixed initially. Then the
2
n.
3
Let V be the volume of the reaction mixture in litres.
number of moles each reacted =
CH3COOH (l) + C2H5OH (l)  CH3COOC2H5(l) + H2O(l)
Initial mole
n
Equilibrium concentration in moles
(n –
Equilibrium concentration
n
2
n)
3
(n –
2
n)
3
0
0
2
n
3
2
n
3
1
n
3
1
n
3
2
n
3
2
n
3
n
3V
n
3V
2n
3V
2n
3V
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MODULE - 5
Chemistry
Chemical Dynamics
[CH 3COOC2 H 5 ][H 2 O]
Kc = [CH COOH][C H OH]
3
2 5
 2n   2n 
   
3V 3V
=
=2×2=4
 n  n 
   
3V 3V
Notes
Kc = 4
Intext Questions 13.2
1.
For a reversible reaction
2A + B  3C + 3D
Write the expression for the equilibrium constant
..............................................................................................................................
2.
What is the relation between Kp and Kc.
..............................................................................................................................
3.
(i) Apply the law of equilibrium to the following and write the expression for K p
and Kc.
(a) CO2 (g) + H2 (g)  CO (g) + H2O (g)
(b) I2 (s)  I2 (g)
(ii) For the above reaction write equation for Kp and Kc.
..............................................................................................................................
4.
The equilibrium constant for the reactions
(i) N2 (g) + 3H2 (g)  2NH3 (g)
1
2
N2(g) + H2(g) 
NH3
3
3
are K1 and K2 respectively. What is the relation between them.
(ii)
..............................................................................................................................
5.
What is the significance of the magnitude of equilibrium constant?
..............................................................................................................................
13.9 Factors Affecting Equilibrium State
Supposing a reaction has reached the equilibrium state and then some conditions like
concentrations, temperature, pressure etc. are changed, would it be affecting the
equilibrium state. If yes how?
242
Chemical Equilibrium
In this section, we shall discuss these questions.
MODULE - 5
Chemical Dynamics
The state of equilibrium is in a dynamic balance between forward and backward reaction.
This balance can be disturbed by changing concentration, temperature or pressure. If
done so a certain net change occurs in the system. The direction of change can be predicted
with the help of Le-Chatelier principle.
13.9.1 Le Chatelier Principles
It states that when a system in equilibrium is disturbed by a change in
concentration, pressure or temperature, a 'net' change occurs in it in a direction
that tends to decrease the disturbing factor.
Notes
The prinicple can be applied to various situations.
13.9.2 Change in Concentration
Consider the state of equilibrium for the formation of ammonia from nitrogen and hydrogen.
N2(g) + 3H2(g) 
 2NH3(g), H = –92.4 kJ/mol
The concentration of nitrogen, hydrogen and ammonia become constant at the point of
equilibrium. Now if any amount of reactants or ammonia is added or removed their
concentration will change and the equilibrium will get disturbed.
(i) Increase concentration of reactant : When the concentration of either nitrogen or
hydrogen is increased; a net forward reaction will take place which consumes the added
reactant.
(ii) Increase in the concentration of any product : If the concentration of product
ammonia is increased, a net backward reaction would take place to utilise the added
ammonia.
13.9.3 Change in Pressure
Change in pressure affects equilibrium involving gaseous phase either in a homogeneous
or hetrogeneons system.
Le Chatelier prinicple for systems involving gases can be studied as follows :
(i) When the number of moles of products is more than the total number of moles of
reactants as in the following system
N2O4(g) 
 2NO2(g)
Increase in total pressure keeping the temperature constant, will cause a decrease in
volume. This means that the number of moles per unit volume will increase. A net change
will take place in the equilibrium in the direction where the number of moles decrease i.e.
backward direction.
(ii) When the number of moles of products is less than reactants. As in the following case
N2(g) + 3H2(g) 
 2NH3(g)
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Chemical Dynamics
Chemistry
According to Le Chatelier's principle increase in total pressure will bring a net change to
the equilibrium in the direction where the total number of moles is decreasing i.e. to the
product side as ng = 2. Decrease in total pressure will bring the net change to equilibrium
in the direction where the total number of moles is increasing i.e. backward direction.
(iii) When there is no change in the total number of moles of reactant and product as in
the following state of equilibrium.
H2(g) + I2(g) 
 2HI
Notes
There is no net change in equilibrium state when pressure is changed.
13.9.4 Change of Temperature
According to Le Chatelier prinicple when the temperature is changed (increased or
decreased) the equilibrium system reacts to nullify the change in heat content. However,
the net change in equilibrium is directed by the exothermic or endothermic nature of
reaction.
(i) Exothermic equilibrium : For the following system of equilibrium of exothermic
nature :
N2(g) + 3H2(g) 
 2NH3(g);
H = – 92.4 kJ/mol
according to Le Chatelier prinicple increase in temperature brings a net change in the
equilibrium state in that direction where this extra heat is consumed. The net change is in
the backward direction and some ammonia will decompose producing nitrogen and
hydrogen. Similarly if the temperature is decreased the equilibrium shifts to the forward
direction.
(ii) Endohermic equilibrium
N2(g) + O2(g) 
H = + 180.7 kJ/mol–1
 2NO(g);
If the temperature is increased the added heat will be absorbed by the reactant and the
net change takes place to the equilibrium in the forward direction. If the temperature in
decreased it will bring a 'net' change to equilibrium in the backward direction i.e. direction
in which it is exothermic.
Addition of a Catalyst : It does not affect the equilibrium. However it helps to achieve
the equilibrium faster.
13.9.5 Applications of Le Chatelier’s Principle
It can be applied to physical as well as chemical equilibria
(A) Physical Equilibria
(1) Melting of Ice
Ice  Water ;
H = + 6 kJ/mol–1
The change of ice to water is endothermic process. According to Le Chatelier principle if
the temperature is increased the net change will take place in the forward direction some
ice will melt into water.
244
Chemical Equilibrium
When the pressure is increased on the equilibrium system, then the volume should decrease;
according to Le Chatelier principle the net change in equilibrium takes place in the forward
direction and ice melts. Therefore, ice melts on increasing the pressure.
MODULE - 5
Chemical Dynamics
(2) Vaporization of Water
Water(l)  Water vapour;
H = + ve
This process occurs with a large increase in volume since ng = 1 – 0 = + 1, and it occurs
with absorption of heat.
Notes
Increasing the temperature results in more vapour formation (endothermic process). Since
ng = + 1, increase in pressure results in a net change in equilibrium in the backward
direction as the volume of water vapours is more than that of liquid water for a given mass
of water.
(3) Solubility Equilibrium
The equilibrium is
Solute (s)  Solute (solution)
The process of dissolution can be endothermic or exothermic. In case of solutes like KCl,
KNO3 and NH4Cl, H is positive (endothermic) and more solute will dissolve on heating.
Thus, the solubility increases with rise in temperature. In case of solutes like KOH and
NaOH the H is negative (exothermic) and their solubility decreases on heating.
(B) Chemical Equilibra
(1) Favourable Conditions for Synthesis of Ammonia : This reaction is of great
industrial importance. During the synthesis of ammonia such conditions are maintained
which favour the ‘net’ forward reaction namely low temperature and high pressure. Addition
of catalyst makes the reaction occur fast. Besides, nitrogen and hydrogen gases are
continuously fed into the reaction chamber and ammonia is continuously removed. All this
keeps the system under stress and equilibrium is never permitted to be attained, so that the
synthesis of ammonia continues to occur.
In industry the reaction is carried out at 4500C and 200 atm pressure in the presence of
finely divided iron (catalyst) and molybdenum (promotor)
(2) Formation of SO3
The reaction
2SO2 (g) + O2 (g)  2SO3 (g) ; H = – ve
is extothermic and ng = 2 – 3 = – 1. Formation of SO3 will be favoured by high pressure
and low temperature in the presence of a catalyst.
(3) Formation of NO
The reaction
N2 (g) + O2 (g)  2NO (g) ;
H = + ve
is endothermic and ng = 2 – 2 = 0. The reaction is not affected by pressure changes and
is favoured at high temperature. Presence of a suitable catalyst would be helpful.
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Chemistry
Chemical Dynamics
Intext Questions 13.3
1.
What is Le Chatelier’ principle?
................................................................................................................................
2.
What are the factors that can affect a system at equilibrium?
................................................................................................................................
Notes
3.
What will happen to solid-vapour equilibrium when the temperature and pressure
are decreased.
................................................................................................................................
4.
(a) Which of the following will result in ‘net’ forward reaction in case of
A (g) + 2B (g)  C (s) + D (g) ; H = + ve
(i) addition of C
(ii) addition of A
(iii) decrease in pressure
(iv) increase in temperature
................................................................................................................................
(b) What are the most favourable conditions for the formation of C and D?
................................................................................................................................
What You Have Learnt
246

A chemical reaction is said to be reversible under certain conditions, if along with the
reactants forming the products, the products also react and form back the reactants
simultaneously.

Reversible reactions do not reach completion stage and result in a state of equilibrium
which is reached when two opposite processes occur at the same rate.

The macroscopic properties of the system do not change once the equilibrium has
been established.

Irreversible reactions are in fact the reversible reactions in which the equilibrium is
reached only when a negligible amount of the reactants is left unreacted.

Chemical equilibrium is dynamic in nature. It can be attained by starting the reaction
from any side and only in a closed system.

When equilibrium is reached as a result of two opposite physical changes, it is called
physical equilibrium and when as a result of two opposite chemical changes it is
called chemical equilibrium.
Chemical Equilibrium

A phase is a homogeneous system or a part of a system which has same composition
and uniform properties throughout. It is not same as physical state.

A system with only one phase is called a homogeneous system and the one with
more than one phases is called heterogeneous system.

Chemical equilibrium can be homogeneous or heterogeneous while physical
equilibrium is always heterogeneous.

For a general reaction aA + bB  cC + dD according to the law of equilibrium, the
equilibrium constant K is given by the expression
MODULE - 5
Chemical Dynamics
Notes
[C]c [D]d
K=
[A]a [B]b

Concentration equilibrium constant Kc is obtained when molar concentration are
used for calculating K. Concentrations of pure solids and liquids are constant and
are not included in the expression of Kc.

In case of gaseous systems, the concentration of gases are expressed in terms of
their partial pressures. The equilibrium constant thus obtained is called the pressure
equilibrium constant, Kp.

The relation between Kp and Kc is = Kc (RT) g where ng is the change in the
number of moles of gaseous substances during the reaction.

Expression of equilibrium constant depends upon how the chemical equation is written
for the reaction.

Magnitude of the equilibrium constant is a measure of how close the reaction is to
the completion stage.

Units of K depends upon the change in the number of moles of the substances
during the reaction.

Concentration, pressure and temperature can affect the equilibrium systems and
the affect can be qualitatively predicted by Le Chatelier’s principle which states
that when a system at equilibrium is disturbed by changing concentration, pressure
or temperature, a ‘net’ change occurs in the direction that tends to neutralize the
effect of the disturbing factor.

Changes in concentration and pressure do result in some chemical reaction, but the
value of the equilibrium constant is not changed.

A catalyst does not change the equilibrium constant. It only helps in reaching the
equilibrium state quicker.

A change in temperature change the value of the equilibrium constant.
n
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MODULE - 5
Chemistry
Chemical Dynamics
Terminal Exercise
1.
What do you understand by reversible and irreversible reactions? Give one example
of each.
...............................................................................................................................
Notes
2.
What is physical equilibrium? Give one example?
...............................................................................................................................
3.
Give characteristics of equilibrium state.
...............................................................................................................................
4.
Is the phase same as physical state? Illustrate your answer with one example of
each.
...............................................................................................................................
5.
How do homogeneous and heterogeneous systems differ from each other? Which
of the following are homogeneous systems?
(a) Liquid  Vapour
(b) N2O4 (g)  2NO2 (g)
(c) NH4Cl (s)  NH3 (g) + HCl (g)
(d) CH3COOH (l) + C2H5OH (l)  CH3COOC2H5 (l) + H2O (l)
...............................................................................................................................
6.
What are Kp and Kc? Derive a relation between them.
...............................................................................................................................
7.
Write down the expression of Kc for the following. Also give units in each case.
(a) N2O5 (g)  2NO2 (g) +
1
O (g)
2 2
(b) CH4 (g) + H2O (l)  CO (g) + 3H2 (g)
(c) FeCl3 (aq) + 3NH4SCN (aq)  Fe (SCN)3(aq) + 3NH4Cl (aq)
...............................................................................................................................
8.
Write down the expression of Kp for the following and give its units (in terms of
atmosphere) in each case
(a) CO2 (g) + H2 (g)  CO (g) + H2O (l)
(b) 3Fe(s) + 4H2O(l)  Fe3O4 (s) + 4H2 (g)
(c) 2SO3 (g)  2SO2 (g) + O2 (g)
248
Chemical Equilibrium
9.
Give the relation between Kc and Kp for the reaction.
MODULE - 5
Chemical Dynamics
CaCO3 (s)  CaO (s) + CO2 (g)
...............................................................................................................................
10. Using the relaction between Kp and Kc write the expression of
(i) Kp for the reactions given in Q. No.7
(ii) Kc for the reactions given in Q. No.8
Notes
...............................................................................................................................
11.
List the factors that can affect
(i) a system at equilibrium and
(ii) equilibrium constant of a system
...............................................................................................................................
12. State the Le Chatelier’s Principle.
...............................................................................................................................
13. What will be the effect of the following factors on the following systems at equilibrium?
2 X (g)  2Y (s) + Z (g); H = + ve
(i) Addition of X,
(ii) removal of Z
(iii) addition of a catalyst
(iv) increasing the pressure and
(v) increasing the temperature.
14. 5 moles of HI were produced by the reaction between 7.5 moles of H2 and 2.6
moles of I2 vapours at 4440C. What is the equilibrium constant of the reaction
H2 (g) + I2 (g)  2HI (g)
15. The equilibrium constant Kp for the reaction
N2O4 (g)  2NO2 (g)
at 333 K is found to be 1.33 atm under a total pressure of 1 atm. Calculate Kp for the
reaction
2NO2 (g)  N2O4 (g)
at 333 K and under 1 atm pressure.
16. At 4440C, 0.30 mole of H2 and 0.30 mole of I2 were taken in a one litre flask. After
some time the equilibrium H2 (g) + I2 (g)  2HI (g) was established and it was
found that the concentration of I2 decreased to 0.06 mol L–1. Calculate the value of
Kc for the reaction at this temperature.
249
MODULE - 5
Chemical Dynamics
Chemistry
17. The equilibrium constant for the reaction.
CH3COOH(l) + C2H5OH (l)  CH3COOC2H5(l) + H2O(l) is 4.0.
What will be the composition of the equilibrium mixture if 1 mole of acetic acid is
taken with 8 moles of ethanol?
18. Kc for the reaction
N2 (g) + 3H2 (g)  2NH3 (g)
Notes
at 4000C was found to be 0.5 L2 mol–2. Calculate Kp of this reaction in atm.
Answers to Intext Questions
13.1
1.
A chemical reaction is said to be reversible, if under certain conditions its products
also react and form back the reactants.
Examples :
H2 (g) + I2 (g)  2HI (g)
2SO2 (g) + O2 (g)  2SO3 (g)
2.
A reaction reaches an equilibrium state when two opposing reactions occur at the
same rate and balance each other at a particular temperature.
3.
When a system reaches the equilibrium state, its temperature, pressure and
concentrations of all the reactants and products do not change any further with time.
4.
(i) Water-vapour system in a closed container at a constant temperature.
(ii) A saturated solution containing some undissolved solute at a constant temperature.
5.
(i) Homogeneous systems :
H2 (g) + I2 (g)  2HI (g)
2SO2 (g) + O2 (g)  2SO3 (g)
(ii) Heterogeneous systems :
CaCO3 (s)  CaO (s) + CO2(g)
Zn (s) + CuSO4 (aq)  Cu (s) + ZnSO4 (aq)
13.2
250
1.
[C]3 [D]3
K = [A]2 [B]
2.
Kp = Kc (RT)
n g
Chemical Equilibrium
3.
pCO  p H2O
[CO][H 2 O]
(i) (a) Kc =
; Kp = p  p
CO2
H2
[CO 2 ][H 2 ]
MODULE - 5
Chemical Dynamics
(b) Kc = [I2]; Kp = PI2
(ii) For the first reaction ng = (1 + 1) – (1 – 1) = 0, hence K c = Kp while for the
second
reaction
ng = 1 – 0 = + 1
 Kp = Kc (RT) or Kc =
4.
Kp
RT
Notes
or Kc < Kp.
[NH 3 ]2
[NH 3 ]2 / 3
K1 =
and K2 =
[N 2 ][H 2 ]3
[N 2 ]1/ 3 [H 2 ]
 K1 = [K2]3.
5.
It is a measure of the extent up to which a reaction proceeds before the equilibrium
is reached.
13.3
1.
Le Chatelier’s principle states that when a system at equilibrium is disturbed by
changing concentration, pressure or temperature, a ‘net’ change occurs in a direction
that tends to neutralize the effect of the disturbing factor.
2.
Changes in pressure, temperature and concentrations of reactants or products.
3.
When the temperature is decreased some vapour will condense and when the pressure
is decreased some solid will sublime.
4.
(a) (ii) and (iv)
(b) High temperature, increase in pressure, presence of a catalyst and continuous
removal of D.
251